A ball is project from the building of height 250m, at angle 45degree below the horizontal with a speed of 25m/s.
i) how long does the ball remain in the air?
ii) find the magnitude and direction of the velocity of the ball just before it strikes the ground. (g=9.81m/s2)
i) Twice the time it takes for the vertical component to become zero.
T = 2*V*sin45/g
ii) The horizontal V component remains constant at Vcos 45 and the vertical component V sin 45 changes sign, but has the same magnitude at the ground, since kinetic energy is the same there.
i) To find how long the ball remains in the air, we can use the vertical motion of the ball. We need to find the time it takes for the ball to reach the ground.
First, let's break down the initial velocity of the ball into its horizontal and vertical components. The initial velocity, V₀, can be split into V₀x (horizontal component) and V₀y (vertical component).
Given:
V₀ = 25 m/s (initial velocity)
θ = 45 degrees (angle below the horizontal)
g = 9.81 m/s² (acceleration due to gravity)
To find V₀x and V₀y, we can use the equations for horizontal and vertical motions:
V₀x = V₀ * cos(θ)
V₀y = V₀ * sin(θ)
Plugging in the values:
V₀x = 25 m/s * cos(45°)
V₀y = 25 m/s * sin(45°)
Now, we can focus on the vertical motion equation:
h = V₀y * t + (1/2) * g * t²
Since the ball starts and ends at the same height, the initial height h is equal to the final height of zero. Thus, we can simplify the equation to:
0 = V₀y * t + (1/2) * g * t²
Rearranging the equation to obtain a quadratic equation in terms of t:
(1/2) * g * t² + V₀y * t = 0
Now, we can solve this quadratic equation for t. Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In the quadratic equation form, we have:
a = (1/2) * g
b = V₀y
c = 0
Substituting the values into the quadratic formula:
t = (-V₀y ± √(V₀y² - 4 * (1/2) * g * 0)) / (2 * (1/2) * g)
Since c is zero, we don't have a term (4ac) in the square root, which simplifies the equation:
t = (-V₀y ± √(V₀y²)) / g
t = (-V₀y ± V₀y) / g
t = 2 * V₀y / g
Plugging in the values:
t = 2 * (25 m/s * sin(45°)) / 9.81 m/s²
Calculating this value should give you the time the ball remains in the air.
ii) To find the magnitude and direction of the velocity of the ball just before it strikes the ground, we can use the vertical and horizontal components of the velocity. The magnitude can be found using the Pythagorean theorem, and the direction can be found using trigonometric functions.
The magnitude of velocity just before it strikes the ground can be calculated using:
V = √(Vx² + Vy²)
Where Vx is the horizontal component of the velocity and Vy is the vertical component of the velocity.
To find Vx, we use the equation Vx = V₀x, since there is no horizontal acceleration:
Vx = V₀x = 25 m/s * cos(45°)
To find Vy, we use the equation for vertical motion:
Vy = V₀y - g * t
Plug in the values:
Vy = (25 m/s * sin(45°)) - (9.81 m/s²) * t
Since we just calculated the time in part i), we can substitute the value:
Vy = (25 m/s * sin(45°)) - (9.81 m/s²) * (time from part i)
Now, we plug both Vx and Vy into the magnitude formula:
V = √(Vx² + Vy²)
Lastly, to find the direction of velocity, we can use the angle of the components:
θ = arctan(Vy / Vx)
Calculating these values should provide you with the magnitude and direction of the velocity just before the ball strikes the ground.